A new Graph Game Problem Description An undirected graph is a graph in which the nodes are connected by undirected arcs. An undirected arc is an edge that has no arrow. Both ends of an undirected arc are equivalent--there is no head or tail. Therefor…

http://acm.hdu.edu.cn/showproblem.php?pid=3435 题意:有n个点和m条边,你可以删去任意条边,使得所有点在一个哈密顿路径上,路径的权值得最小. 思路: 费用流,注意判断重边,否则会超时. #include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<queue> using namespace std; ty…

题意: 有S1到Sn这n个勇士要和X1到Xn这n个勇士决斗,初始时,Si的决斗对象是Xi. 如果Si赢了Xi,那么你将获得Vi分,否则你将获得-Vi分. Si和Xi对决时,Si有初始生命Hi,初始攻击Ai, Xi有初始生命Pi,初始攻击Bi. 且Si先出手,然后Xi失去Ai生命,之后如果Xi没死,那么Xi出手,Si失去Bi生命. 直到有一方的生命值<=0时,决斗结束. 现在要你重新安排S和X的决斗顺序,使得你能获得的分最多.如果有多个最优解,你要选取那个维持初始决斗顺序最多的解. 解析: 仔细看…

Problem Description An undirected graph is a graph in which the nodes are connected by undirected arcs. An undirected arc is an edge that has no arrow. Both ends of an undirected arc are equivalent--there is no head or tail. Therefore, we represent a…

http://acm.hdu.edu.cn/showproblem.php?pid=2485 题意: 现在要从起点1到终点n,途中有多个车站,每经过一个车站为1时间,现在要在k时间内到达终点,问至少要破坏多少个车站. 思路: 把每个点拆分为两个点,容量为1,费用为0.之后相邻的车站连边,容量为INF,费用为1,表示经过一个车站需要1时间. 这样一来,跑一遍费用流计算出在费用不大于k的情况下的最大流,也就是最小割,即至少要破坏的车站数. 在网络中寻求关于f的最小费用增广路,就等价于在伴随网络中寻求…

HDU 2686 Matrix 题目链接 3376 Matrix Again 题目链接 题意:这两题是一样的,仅仅是数据范围不一样,都是一个矩阵,从左上角走到右下角在从右下角走到左上角能得到最大价值 思路:拆点.建图,然后跑费用流就可以,只是HDU3376这题,极限情况是300W条边,然后卡时间过了2333 代码: #include <cstdio> #include <cstring> #include <vector> #include <queue>…

题意: 有\(n\)个数\(a_1\cdots a_n\),现要你给出\(k\)个不相交的非降子序列,使得和最大. 思路: 费用流建图,每个点拆点,费用为\(-a[i]\),然后和源点连边,和后面非降的数连边,源点和超级源点连一条容量\(k\)的边,跑费用流. 用\(spfa\)费用流\(TLE\),这里因为不会出现负环,所以用\(Dijkstra\)优化. 代码: /******* dijkstra优化费用流模板 *******/ //不能有负环 #include<functional> /…

A new Graph Game Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1849    Accepted Submission(s): 802 Problem Description An undirected graph is a graph in which the nodes are connected by undir…

http://acm.hdu.edu.cn/showproblem.php?pid=3435 #include <cstdio> #include <iostream> #include <cstring> #include <queue> #include <algorithm> #define inf 0x3f3f3f3f #define maxn 54444 using namespace std; queue<int>q; s…

度度熊的交易计划 Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1192    Accepted Submission(s): 440 Problem Description 度度熊参与了喵哈哈村的商业大会,但是这次商业大会遇到了一个难题: 喵哈哈村以及周围的村庄可以看做是一共由n个片区,m条公路组成的地区. 由于生产能力的区别,第i…